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WJEC 2014 Online Exam Review

GCSE Additional Science Physics Higher 4473-02

All Candidates' performance across questions

Question Title N Mean S D Max Mark F F Attempt %1 4884 8.7 1.8 12 72.5 1002 4876 6.2 2.5 12 51.5 99.83 4881 4.7 2.6 10 47 99.94 4879 4.1 2.4 12 34.2 99.95 4882 7.5 2.7 14 53.6 100

72.5

51.5

47

34.2

53.6

0 10 20 30 40 50 60 70 80 90 100

1

2

3

4

5

Facility Factor %

Que

stio

n

GCSE AdditionalScience Physics Higher 4473-02

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Usually the question number

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The number of candidates attempting that question

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The mean score is calculated by adding up the individual candidate scores and dividing by the total number of candidates. If all candidates perform well on a particular item, the mean score will be close to the maximum mark. Conversely, if candidates as a whole perform poorly on the item there will be a large difference between the mean score and the maximum mark. A simple comparison of the mean marks will identify those items that contribute significantly to the overall performance of the candidates. However, because the maximum mark may not be the same for each item, a comparison of the means provides only a partial indication of candidate performance. Equal means does not necessarily imply equal performance. For questions with different maximum marks, the facility factor should be used to compare performance.

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The standard deviation measures the spread of the data about the mean score. The larger the standard deviation is, the more dispersed (or less consistent) the candidate performances are for that item. An increase in the standard deviation points to increased diversity amongst candidates, or to a more discriminating paper, as the marks are more dispersed about the centre. By contrast a decrease in the standard deviation would suggest more homogeneity amongst the candidates, or a less discriminating paper, as candidate marks are more clustered about the centre.

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This is the maximum mark for a particular question

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The facility factor for an item expresses the mean mark as a percentage of the maximum mark (Max. Mark) and is a measure of the accessibility of the item. If the mean mark obtained by candidates is close to the maximum mark, the facility factor will be close to 100 per cent and the item would be considered to be very accessible. If on the other hand the mean mark is low when compared with the maximum score, the facility factor will be small and the item considered less accessible to candidates.

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For each item the table shows the number (N) and percentage of candidates who attempted the question. When comparing items on this measure it is important to consider the order in which the items appear on the paper. If the total time available for a paper is limited, there is the possibility of some candidates running out of time. This may result in those items towards the end of the paper having a deflated figure on this measure. If the time allocated to the paper is not considered to be a significant factor, a low percentage may indicate issues of accessibility. Where candidates have a choice of question the statistics evidence candidate preferences, but will also be influenced by the teaching policy within centres.

8

(4473-02)

Examineronly

3. Living animals take in small amounts of radioactive carbon-14. After death, the amount of carbon-14 in their bodies decreases, because the carbon-14 atoms decay. The amount of carbon-14 remaining in the bones of an animal’s skeleton can be used to estimate its age.

Carbon-14 is a beta emitter, with a half-life of 5 720 years.

(a) State what is meant by the following statements: [3]

(i) carbon-14 is a beta emitter;

(ii) carbon-14 has a half-life of 5 720 years.

(b) Complete the decay equation for carbon-14 shown below. [3]

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(c) (i) A bone taken from a skeleton, found at an archaeological site, contains 10 units of carbon-14. An identical bone in a living animal contains 160 units of carbon-14. Use your understanding of half-life to calculate the age of the skeleton. [2]

age = . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . years

(ii) Explain why this method of calculating the age of bones is unreliable for skeletons believed to be less than 100 years old. [2]

10

614C N e+. . . . . .

. . . . . .

. . . . . .

. . . . . .

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2 marks - the method showed 4 half-lives and the answer is correct.

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3 marks - full marks awarded here.

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3 marks - (i) this statement was acceptable. (ii) the reference to the Becquerel was accepted as a reference to activity.

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1 mark - this statement gained the first mark. The remainder of the response is not an appropriate linked statement.

NYDW

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Question total = 9

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Question total = 5.

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1 mark - this is an equivalent form of words to that shown in the marking scheme as the linked statement for the second mark. It can still be awarded a mark, even though the first statement is omitted.

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1 mark - the method shows 4 half-lives. Use of this value is incorrect. Only the first mark is awarded.

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0 marks - all values are incorrect.

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3 marks - (i) perfect answer. (ii) the time for a property to halve is included. The addition of the phrase 'count rate' resulted in the second mark being awarded.

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Question total = 8.

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0 marks - the answer is too vague. It is too much of an assumption that the phrase 'a small number' refers to 100 years which is a (small) fraction of the half life.

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2 marks - suitable method showing 4 half-lives and a correct calculation of age.

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3 marks - all values correct.

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3 marks - (i) perfect answer. (ii) the statement about a property halving in a time is included for the first mark. The number of C-14 nuclei is an acceptable property for the second mark.

(4473-02) Turn over.

9Examiner

only4. On 14 October 2012 Felix Baumgartner created a new world record when he jumped from a

stationary balloon at a height of 39 km above the surface of the Earth. At 42 s of free fall he reached a terminal velocity of 373 m/s, which was greater than the speed of sound. The total mass of Felix and his suit was 118 kg.

(a) Explain in terms of weight and air resistance how terminal velocity is reached. [3]

(b) (i) Use an equation from page 2 to calculate Felix’s change in momentum in the first 42 s of his fall. [2]

change in momentum = . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . kg m/s

(ii) Use an equation from page 2 to calculate the mean resultant force acting on him during the first 42 s. [2]

mean resultant force = . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . N

(iii) Calculate the mean value of the air resistance force during the first 42 s. [3]

mean air resistance force = . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . N

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(4473-02)

10

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(c) At 39 km the air particles are very far apart. Explain how jumping from this height allowed Felix to reach such a high terminal velocity. [2]

Examineronly

12

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0 marks - this method just resulted in finding the force already calculated in (ii). No attempt to calculate the weight and then the difference between the two forces.

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2 marks - correct method and calculation.

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2 marks - this was accepted as an alternative method of calculating the change in momentum.

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1 mark - this answer only referred to balanced forces at terminal velocity without any explanation of why they become balanced. The third marking point was awarded. The statement that weight is 118 kg did not attract a penalty.

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This equation is not mathematically correct but did not attract a penalty.

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Question total = 7.

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2 marks - both marking points present and they are linked as one being the consequence of the other.

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0 marks - this value is given in the question stem at the top of the page, and is actually the terminal velocity.

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0 marks - this candidate substituted a momentum value as the acceleration. It was quite common for the previous answer to be used incorrectly in a subsequent part of the question such as this.

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2 marks - correct method and answer. The initial momentum being zero is not mentioned but from the value on the answer line it is assumed this was taken into account.

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1 mark - this answer only referred to balanced forces at terminal velocity without any explanation of why they become balanced. The third marking point was awarded.

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Question total = 4.

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1 mark - this is given in the marking scheme as one of the alternative statements for the second mark. It can be awarded a mark even though the first statement is omitted.

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3 marks - this is a correct solution. A force has been calculated from rate of change of momentum and the value of the weight is shown. The question arises whether this should be treated independently from part (ii) or follow the error through. In this instance it is given full credit.

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0 marks - this is a calculation of weight. The candidate crossed out the correct method.

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2 marks - correct method and answer. The initial momentum being zero is not mentioned but from the value on the answer line it is assumed this was taken into account.

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1 mark - the first sentence is incorrect. This answer goes on to refer to balanced forces at terminal velocity without any explanation of why they become balanced. The third marking point was awarded.

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Question total = 8.

NYDW

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2 marks awarded.

11

(4473-02)

Examineronly

5. (a) Describe how you would investigate how the resistance of a filament lamp changes with the voltage. [6 QWC]

Include in your answer:

• a labelled circuit diagram;

• how you would obtain a range of results;

•howyouwouldanalysethedata.

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TURN OVER FOR THE REST OF THE QUESTION.

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Question total = 3.

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3 marks - the resistance equation is included. The account about voltage is confused. The statement about current is incorrect. The structure of some sentences is clumsy. This is worthy of a middle band mark, and 3 marks are awarded.

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The circuit will work. The variable voltage supply is labelled. The gap between the cell terminals is not penalised.

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Question total = 1.

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1 mark - the task was misunderstood. The candidate thinks that there will be a difference in the ammeter readings in the series circuit. However, the circuit would work and the current could be measured. The response is considered to be lower band and 1 mark is given.

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The circuit would work if connected. However, it lacks all the required components to carry out the task.

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Question total = 5.

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5 marks - the candidate describes how the voltage will be varied although it is an awkward method. The measurements that will be taken are included. The equation to calculate resistance is also included. The accuracy of the account is fine. This is a top band answer and is awarded 5 marks.

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The circuit lacks a variable resistor, but otherwise it is fine.

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17

Question Marking details Marks

3. (a) (i) Emits an electron 1

(ii) It is the time taken to halve / it takes 5 720 years to halve(1) the

number of C-14 nuclei (or atoms) / the mass (or amount) of C-14 / the

activity / count rate (1) Treat radioactivity as neutral.

2

(b) (2) (1) beware of missing minus sign 3

(c) (i) It takes 4 half-lives (1) so time = 4 × 5 720 = 22 880 [years] (1) 2

(ii) 100 years is only a [small] fraction or 157

of a half-life (1) so difference

in C-14 readings will be small / won’t have decayed by very much (1)

To award both marks both statements must be linked.

2

Question total [10]

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17


4. (a) Initially weight is greater than air resistance [so he accelerates] (1) as

he goes faster air resistance increases (1) eventually the forces balance

(1) To award all 3 marks the third statement written down must be

linked to one of the previous two statements.

3

(b) (i) Change in momentum = 118 × 373 (1) = 44 014 [kg m/s] (1) 2

(ii) Resultant force =

44014

42(1) = 1 048 [N] (1) ecf from (i)

OR use of F = ma = 118 × 8.9 (1) = 1 050 [N] (1)

2

(iii) W = 118 × 10 = 1 180 N (1)

Air resistance = W – resultant force or 1 180 – 1 048 (1)

= 132 [N] (1) ecf from (ii) & on weight

N.B. Answer mark awarded only if correct sign present

e.g. 118 – 1 048 (1) = - 930 [N]

3

(c) Air resistance force small (1) because in contact with small number of

air particles [per second] / so need to go very quickly for air resistance

to balance weight / so need to go very quickly to reach terminal

velocity (1) To award both marks both statements must be linked.

2

Question total [12]

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18


5. (a) Indicative content:

6

Diagram showing the lamp connected in series with an ammeter, a

variable power supply or a power supply with a variable resistor. A

voltmeter is connected in parallel across the lamp. A pair of readings is

taken from the ammeter and voltmeter. The variable resistor is

adjusted [to vary the voltage across the lamp] and another pair of

readings is taken. This is repeated. Results are plotted on a graph and

the variation in resistance can be determined by how the slope changes

OR the resistance of the lamp is calculated for each set of readings by

using V=IR.

5 – 6 marks The candidate constructs an articulate, integrated account

correctly linking relevant points, such as those in the indicative

content, which shows sequential reasoning. The answer fully addresses

the question with no irrelevant inclusions or significant omissions. The

candidate uses appropriate scientific terminology and accurate

spelling, punctuation and grammar.

3 – 4 marks The candidate constructs an account correctly linking

some relevant points, such as those in the indicative content, showing

some reasoning. The answer addresses the question with some

omissions. The candidate uses mainly appropriate scientific

terminology and some accurate spelling, punctuation and grammar.

1 – 2 marks The candidate makes some relevant points, such as those

in the indicative content, showing limited reasoning. The answer

addresses the question with significant omissions. The candidate uses

limited scientific terminology and inaccuracies in spelling, punctuation

and grammar.

0 marks The candidate does not make any attempt or give a relevant

answer worthy of credit.

(b) (i) Scales (at least ½ of each axis used) (1) plots (allow ± ½ small square

division) (1) smooth curve allow ± ½ small square division (1) allow

straight line between 0 and 2 and between 2 and 6 but expect a curve

between 6 and 10. Don’t allow wispy, wobbly, thick lines

3

(ii) Current from their graph (i.e. 1.6 A) (1), substitution (1),

answer = 3.1 [] (1) ecf

3

(iii) R must be increasing (1) Graph becomes less steep / as voltage

increases, current increases less and less / an alternative method would

be calculating another value of R (1)

To award both marks both statements must be linked.

2

Question total [14]

Higher tier paper total [60]

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